2 Answers
2

Let $H_X, H_Y,T_X,T_Y$ denote the heads and tails counts of $X$ and $Y$.
Assume that the heads of $X$'s coin is red, tails green, whereas the heads of $Y$'s coin is green, tails red.
We ask for the probability of
$$H_Y>H_X$$
$$\iff T_X+H_Y=n-H_X+H_Y>n+H_X-H_Y=H_X+T_Y-1\\\iff T_X+H_Y\ge T_Y+H_X$$
But the latter expression is just that the number of grean outcomes is at least as large as the number of red ones.
For this, the answer (by symmetry and since ties cannot occur) $\frac12$.

Alternatively, let $Y$ delay her last throw.
With both players making $n$ throws, there is a certein probability $p$ that $X$ has more heads, the same probaility that $Y$ has more heads, and the probability $1-2p$ for a tie.
With the last coin, $Y$ can only make a change in case of a tie and does so half the time for a win. That is: The probability of more heads for $Y$ is $p+\frac{1-2p}2=\frac12$.

Let $X_n$ and $Y_n$ be the number of head in the first $n$ tosses. What is asked for is the probability of
$$P(X_n < Y_{n+1}) = P(X_n < Y_n) + P(X_n = Y_n)P(\text{last toss of }Y \text{is a head})$$
Apparently, $P(X_n<Y_n)+P(X_n>Y_n)+P(X_n=Y_n)=1$ and because of symmetry, $P(X_n<Y_n)=P(X_n>Y_n)$ so that $P(X_n<Y_n)=\frac{1-P(X_n=Y_n)}{2}$. Moreover, $P(\text{last toss of }Y \text{is a head})=1/2$. Plug them into the above equation, you will find out that $$P(X_n<Y_{n+1})=\frac{1}{2}$$